A plane electromagnetic wave of frequency 35 | vedclass

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Question

(A) The relationship between the electric field $E$ and the magnetic field $B$ in an electromagnetic wave is given by $E/B = c$,where $c$ is the speed

Vedclass · Accepted Answer

$3.2 	imes 10^{-8} \ \hat{k} \ T$

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(A) The relationship between the electric field $E$ and the magnetic field $B$ in an electromagnetic wave is given by $E/B = c$,where $c$ is the speed of light in free space $(c = 3 	imes 10^8 \ m/s)$.First,calculate the magnitude of the magnetic field:$B = E/c = 9.6 / (3 	imes 10^8) = 3.2 	imes 10^{-8} \ T$.Next,determine the direction of the magnetic field using the property that the direction of propagation $\hat{v}$ is given by $\hat{v} = \hat{E} 	imes \hat{B}$.Rearranging this,the direction of the magnetic field is $\hat{B} = \hat{v} 	imes \hat{E}$.Given the wave travels in the $X$-direction $(\hat{v} = \hat{i})$ and the electric field is in the $Y$-direction $(\hat{E} = \hat{j})$:$\hat{B} = \hat{i} 	imes \hat{j} = \hat{k}$.Therefore,the magnetic field vector is $\overrightarrow{B} = 3.2 	imes 10^{-8} \ \hat{k} \ T$.

$A$ plane electromagnetic wave of frequency $35 \ MHz$ travels in free space along the $X$-direction. At a particular point (in space and time) $\overrightarrow{E} = 9.6 \ \hat{j} \ V/m$. The value of the magnetic field at this point is:

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